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ck99
Nov3-11, 05:50 AM
Through my mathematical fumblings, I think I have found a metric which gives a solution of the geodesic equation of motion that is asymptotic. It is a diagonal metric, with g00 = (x_1)^(-3) and g11 = 1. I am largely self-taught with SR so I may be miles off, but I think this gives a G.E. of M which has a λ = 1 / (x_1) term in it, so my parameter goes to infinity when x_1 = 0.

Even if I have the details wrong, I have two questions:

1) Can you define a metric where geodesics are asymptotic?

2) Can you define the same metric with different coordinates to remove this behaviour?

The only thing I can think of is some kind of substitution, but I don't really know what to do and the textbook I am working through is not a lot of help. I even got the massive "Gravitation" book out of my library but if it did contain the solution, I didn't understand it!

Hopefully someone can give me a clue :)

Bill_K
Nov4-11, 03:18 PM
In the Schwarzschild metric in the usual coordinates, the inward-going null geodesics run off the coordinate patch, going to t = +∞ as r goes to 2m. Is that what you mean by asymptotic?

Matterwave
Nov4-11, 06:05 PM
In the Schwarzschild metric in the usual coordinates, the inward-going null geodesics run off the coordinate patch, going to t = +∞ as r goes to 2m. Is that what you mean by asymptotic?

If this is what you meant by asymptotic, then whether you can get rid of this behavior via a change of coordinates is dependent on if this behavior is a real singularity (e.g. at the center of a black hole) or a coordinate singularity (e.g. at the event horizon of a black hole). A real singularity would have the curvature tensor also diverging, whereas a coordinate singularity would have a finite curvature. If it's a coordinate singularity then you can find coordinates which remove this asymptotic behavior (for Schwarzschild, you can use Eddington-Finklestein coordinates or Kruskal-Szekeres coordinates).